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Let $(M,g)$ be a Riemannian manifold. The musical isomorphisms $^\flat:\chi(M) \to \Omega^1(M)$ and $^\sharp:\Omega^1(M) \to \chi(M)$ allow the space of differential one-forms $\Omega^1(M)$ to be identified with the space of vector fields $\chi(M)$.

If I'm not mistaken, I can define the Lie bracket of two differential one forms $\alpha,\beta$ by $$[\alpha,\beta] := [\alpha^\sharp, \beta^\sharp]^\flat.$$

Now suppose that $\alpha$ and $\beta$ are exact; i.e. there exist smooth functions $A,B:M\to \mathbb{R}$ such that $\alpha = dA$ and $\beta = dB$, where $``d"$ denotes the exterior derivative. Is it necessarily true that $[dA,dB] = 0$?

Here is the motivation for my question: On a $k$-manifold $M$, the integral curves of $k$ vector fields linearly independent at the point $x \in M$ are the coordinate curves of a local coordinate system centered at $x$ if and only if their pairwise Lie brackets are zero (see, e.g., the discussion here). On a Riemannian manifold, differential one-forms also have integral curves after identifying these one-forms with vector fields. I would like to know if there are "natural" conditions on a collection of $k$ differential one-forms that determine whether their integral curves similarly form the coordinate lines of a coordinate chart.

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    $\begingroup$ This is just the Lie bracket of gradient vector fields. Try some examples on $\mathbb R^2$. $\endgroup$ – Anthony Carapetis Mar 16 '16 at 4:27
  • $\begingroup$ I should have done this from the beginning. Thank you. $\endgroup$ – Matthew Kvalheim Mar 16 '16 at 11:43
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Answer: The answer to my question is that $[dA,dB]$ is not necessarily zero. As a counterexample, consider $\mathbb{R}^2$ with the standard inner product and take $A(x,y) = x^2y^2$, $B(x,y) = xy^3$ (I selected these arbitrarily without thinking about them). Then $\nabla A(x,y) = (dA)^\sharp(x,y) = 2xy^2\frac{\partial}{\partial x} + 2x^2y \frac{\partial}{\partial y}$ and $\nabla B(x,y) = (dB)^\sharp(x,y) = y^3 \frac{\partial}{\partial x} + 3xy^2\frac{\partial}{\partial y}$.

Computing the Lie Bracket of $\nabla A$ and $\nabla B$ in coordinates shows that $[dA,dB](x,y) = [\nabla A, \nabla B](x,y) = (-6x^2y^3 - 2y^5)\frac{\partial}{\partial x} + (2xy^4 + 6x^3y^2)\frac{\partial}{\partial y}$ which is not zero at all points.

Resolving my misunderstanding: The following is an explanation of the misunderstanding which prompted my question. Let $A_1,\ldots,A_k$ be functions on a Riemannian $k$-manifold $(M,g)$ such that the collection of vectors $\{\nabla A_i := (dA_i)^\sharp\}$ are linearly independent at the point $x_0 \in M$. This is true if and only if the collection of forms $\{dA_1,\ldots,dA_k\}$ are themselves linearly independent as linear functionals at $x_0 \in M$. Then the derivative of the map $\varphi: M \to \mathbb{R}^k$, $\varphi: x \mapsto (A_1(x),\ldots,A_k(x))$ is full rank at $x_0 \in M$, so the inverse function theorem guarantees that there exists an open set $U \ni x$ such that $\varphi|_U: U \to \varphi(U)$ is a diffeomorphism. Initially, I thought that the pairwise Lie brackets of all of the $\nabla A_i$ must therefore be zero. However, this is only necessarily be true if the integral curves of the $\nabla A_i$ were the coordinate lines determined by the chart $\varphi$. The definition of $\nabla A_i$ relies on the particular metric $g$, and in general $g$ has nothing to do with $\varphi$; the coordinate lines of $\varphi$ which have nothing do with $g$ are thus not the integral curves of the $\nabla A_i$ in general.

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